If ab ≠ 0, is |a-b| > |a+b|?

If \(ab ≠ 0\), is \(|a-b| > |a+b|\)?

Square \(|a-b| > |a+b|\) (we can safely do this since both sides are nonnegative):

Is \(a^2 - 2ab + b^2 > a^2 +2ab + b^2\)?
Is \(ab < 0\)?

(1) \(ab < 0\). Directly answers the question. Sufficient.

(2) \(a > b\). Not sufficient to say whether ab < 0.

Answer: A.
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Hi,
Ofcourse the above method by Bunuel is the right and best method, But what it means logically..

is \(|a-b| > |a+b|\)?..
We have a number line in which a and b and -b are marked..
the MOD asks us whether a is farther from b than -b..

two cases-

A) if both are same sign, a will be closer to b than -b, since midpoint of b and -b will be 0.. ans will be NO
B) if both are opposite sign, a will be closer to -b and ans will be YES..

1) \(ab < 0\)
Both a and b are of opposite sign .. ans YES
Suff

2) \(a > b\)
Nothing ..
both can be same sign or opposite sign ..
the above TWO cases still remain..
Insuff
_________________

For |a-b| > |a+b|, a and b should have opposite signs. So, if a > 0, b should be less than 0 and vice versa.

1) ab < 0
this proves a and b have opposite signs. So, sufficient.

2) a>b
not sufficient as it does not prove a and b will have opposite signs.
for e.g. a=1, b=-2 --> |3| > |1| --> Yes
a=3, b =2 --> |1| < |5| --> No

Thanks for the reply, Bunuel! Just a quick follow-up: can we assume that both sides are non-negative since ab ≠ 0?

Both sides are nonnegative because both sides are absolute values which cannot be negative.
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Hope this helps

Hi chetan2u,

I tried other approach by opening the modulus but it seems there is an error.

is |a−b|>|a+b|?

case 1:

a-b > a+b……… 2b<0……..so b<0.. B is negative

case 2:

a-b < -a-b …….. 2a<0……..so a< 0 ..a is negative


Statement 1:

ab<0

According to my analysis above, it should be insufficient because both a & b should negative.


Statement 2:

a>b It should be insufficient


Can you help with identifying the problem with above solution?

Thanks in advance

Hi,

the problem in your solution is opening of MOD..
yes, you can save a lot of time in opening of MOD, if both sides are +ive but do not open it straightway, SQUARE it..

Do not waste time in taking four scenarios, but square both sides, it will siffice..

\(|a-b|>|a+b|..\)
square both sides..
\((a-b)^2 > (a+b)^2...................\\
a^2+b^2-2ab>a^2+b^2+2ab............\\
4ab<0....\)....
so we should find out if ab<0

and statement I tells us that ab<0 .. so suff
_________________

Thanks Z10201Z for your help


Hi chetan2u,

Thanks you for your quick response. I certainly will follow the rule in case of two modulus with inequality.

Aster reading your post about 'Absolute modulus : A better understanding', I have two questions:

1- I want to take further to solve a confusion. I have seen your way solving in one problem when two modulus in equality: |X|=|Y|

X=Y or X=-Y or -X=-Y (which is same as X=Y) or -X=Y (Which is the same as X=-Y). However, the previous technique is valid only for Equity and invalid for inequality. Am I right? I think this my error.


2- if the question is |a-b|>a+b ............what the best to solve it? In the case, I'm curious to know how to open modulus just to get to know.


I hope one day you can make another post for absolute modulus under inequality.

Thanks in advance for your help. I appreciate it.

hi
lets solve |a-b|>a+b..
this can be straight ..
1) a-b>a+b......2b<0..... b<0..
2) -(a-b)>a+b......-a+b>a+b.....2a<0..... a<0..

Or
Critical points..

a is 0, when a is b, so b is our critical point..
this gives us two ranges
a<b or a>=b..
1) a<b...
then a-b will be -ive..
so -(a-b)>a+b......-a+b>a+b.....2a<0..... a<0..
2) a>=b..
then a-b will be +ive or 0..
so a-b>a+b......2b<0..... b<0..

Best METHOD for any MOD irrespective of = or inequality, follow "critical point" method
_________________

You can think of absolute value as distance from zero. So we need enough info to definitely determine whether a - b is further from zero than the sum a + b.

Statement one tells us that a and b have different signs (that's the only way that the product will be negative).

For the difference a - b, this means that the numbers end up 'working together' to get away from zero; if you take a positive number away from a negative number, you go further down, further away from zero (e.g. -10 - 2 = -12). If you take a negative number away from a positive number you go further up (e.g. 10 - - 2 = 12) Takeaway: The difference of numbers with opposite signs is always further from zero than either of the original numbers.

For the sum a + b, statement one means that the numbers end up 'working in opposite directions.' That is, when we add numbers of opposite signs, one cancels out part of the other's distance to zero. Think of 10 + - 2 and -2 + 10. In both cases we end up closer to zero than the furthest term was to begin with. Takeaway: the sum of two numbers with opposite signs will be closer to zero than the original number of greatest magnitude (the term that was furthest from zero to begin with).

So if we know two numbers have opposite signs, we know the difference of those numbers ends up further from zero than the sum. Sufficient.


statement 2 doesn't tell us anything about the signs, so we might be suspicious right from the start. And sure enough, if all we know is that a is to the right of b, then we might have a couple of positives, (such as a = 3, and b = 2) in which case the sum is further from zero than the difference. But we could also have a positive and a negative (such as a = 3, and b = - 2), in which case the difference is further from zero than the sum. Insufficient.

1) ab < 0

Means that there are two possibilities: either a>0 and b<0 or a<0 and b>0. Let's look at htose two seperately.

a>0, b<0. Let's denote b=-y (y - some positive integer). -y<0, y>0.

|(a - (-y))| > |a - y|

|a + y| > |a - y|

On the left side we have sum of two positive integers so we can pull off our modulus. On the right side we have difference o two positive integers which will always be lesser than their sum.

a<0, b>0. Same approach. a=-x (x some positive integer). -x<0, x>0

|-x - b| > |-x + b|

| - (x +b)| > |b - x|

x + b > |b - x|

Again we have sum of two positive integers on the left and their diffrence on th right. Sufficient.

2) a > b

We have three possible scenarious.

a) a>0, b>0
b) a<0, b<0
c) a>0, b<0

a) To eveade confusion it would be impler to use particular numbers here.

a = 30, b = 10

|30 - 10| ... |30 + 10| ----> 20 < 40

b) a = -10 b = -30

|-10 - (-30)| ... | -10 - 30|

| -10 + 30| ... | - (10 + 30)|

20 < 40

c) In c we have same case as in (1)

a=10, b=-30

|10 - (-30)| ... |10 - 30|

40 > 20 Insufficient

Answer A

I hope my approach wasn't too vague. May be our experts will give you more details.

Learn More about Absolute Values Here

|a-b|>|a+b| will be true only when a and b have different signs, as when we are doing
-> |a-b| we are actually adding the values of a and b and
-> |a+b| we are actually subtracting the values of a and b.

STAT1
ab < 0 => a and b have different signs
SUFFICIENT

STAT2
a > b
a and b can have the same sign or can have different signs too
example:
1. a = 7 b = 2 => a> b
but |a-b|<|a+b| (|5| < |9|)
2. a = 7 b = -2 => a> b
|a-b|>|a+b| (|9| > |5|)
So, INSUFFICIENT

So, Answer will be A
Hope it helps!

Watch the following video to learn the Basics of Absolute Values


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Hi Bunuel, to confirm, the sign of the sides matters because if we multiply by a -ve, then the inequality sign would need to flip, correct?

We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality).
For example:
\(2<4\) --> we can square both sides and write: \(2^2<4^2\);
\(0\leq{x}<{y}\) --> we can square both sides and write: \(x^2<y^2\);

But if either of side is negative then raising to even power doesn't always work.
For example: \(1>-2\) if we square we'll get \(1>4\) which is not right. So if given that \(x>y\) then we cannot square both sides and write \(x^2>y^2\) if we are not certain that both \(x\) and \(y\) are non-negative.

Check the link below for more:
Inequality tips
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Hai,
I am new here. I have a doubt. Where are the options mentioned, I don't see any A,B,C,D and E beneath the questions.

This is a data sufficiency question. Options for DS questions are always the same.

The data sufficiency problem consists of a question and two statements, labeled (1) and (2), in which certain data are given. You have to decide whether the data given in the statements are sufficient for answering the question. Using the data given in the statements, plus your knowledge of mathematics and everyday facts (such as the number of days in July or the meaning of the word counterclockwise), you must indicate whether—

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked.
C. BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked.
D. EACH statement ALONE is sufficient to answer the question asked.
E. Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed.

I suggest you to go through the following post ALL YOU NEED FOR QUANT.

Hope this helps.
_________________

Similar Questions to practice:

https://gmatclub.com/forum/if-ab-0-is-a ... 17927.html
https://gmatclub.com/forum/if-ab-ab-is- ... l#p1900871
https://gmatclub.com/forum/is-a-b-a-b-2 ... l#p1900872
https://gmatclub.com/forum/is-a-b-a-b-246248.html
https://gmatclub.com/forum/is-a-b-a-b-246249.html

The best way to solve this is by using Bunuel's method. The next best way is to use smart numbers and plug them into the various equations.

However, if you don't see that Bunuel's method is possible during the ~ 2min you have to solve the problem or if your smart numbers aren't working, then the following method is a pretty standard way to arrive at the right solution. Keep in mind though that this method does take longer and should be used as a kind of last resort.

\(\lvert a - b\rvert > \lvert a + b\rvert\)

\(\lvert a - b\rvert = a - b\), when \(a - b > 0\)
\(\lvert a - b\rvert = -(a - b)\), when \(a - b < 0\)
\(\lvert a + b\rvert = a + b\), when \(a + b > 0\)
\(\lvert a + b\rvert = -(a + b)\), when \(a + b < 0\)

Case 1:
if \(a - b > 0, a + b > 0\), then
\(a - b > a + b\)
\(0 > 2b\)
\(b < 0\)
Also, if we add \(a + b > 0\) and \(a - b > 0\) we get:
\(a > 0\)

Case 2:
if \(a - b > 0, a + b < 0\), then
\(a - b > -(a + b)\)
\(a - b > -a - b\)
\(2a > 0\)
\(a > 0\)
Also, if we subtract \(a + b < 0\) from \(a - b > 0\) we get:
\(b < 0\)

Case 3:
if \(a - b < 0, a + b > 0\), then
\(-(a - b) > a + b\)
\(-a + b > a + b\)
\(0 > 2a\)
\(a < 0\)
Also, if we subtract \(a + b > 0\) from \(a - b < 0\) we get:
\(b > 0\)

Case 4:
if \(a - b < 0, a + b < 0\), then
\(-(a - b) > -(a + b)\)
\(-a + b > -a - b\)
\(2b > 0\)
\(b > 0\)
Also, if we add \(a + b < 0\) and \(a - b < 0\) we get:
\(a < 0\)


For all of our cases, if \(a < 0\), then \(b > 0\) or if \(a > 0\), then \(b < 0\).
This is the same as \(ab < 0\).

More on this method here and here.